Codes Correcting Two Deletions

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1 1 Code Correcting Two Deletion Ryan Gabry and Frederic Sala Spawar Sytem Center Univerity of California, Lo Angele Abtract In thi work, we invetigate the problem of contructing code capable of correcting two deletion. In particular, we contruct a code that require redundancy approximately 8 log 2 n + O(log 2 log 2 n) bit of redundancy, where n denote the length of the code. To the bet of the author knowledge, thi repreent the bet known contruction in that it require the lowet number of redundant bit for a code correcting two deletion. arxiv: v1 [c.it] 19 Dec 2017 I. INTRODUCTION Thi paper i concerned with deletion-correcting code. The problem of creating error-correcting code that correct one or more deletion (or inertion) ha a long hitory, dating back to the early 1960 [12]. The eminal work in thi area i by Levenhtein, who howed in [9] that the Varhamov-Tenengolt aymmetric error-correcting code (introduced in [13]) alo correct a ingle deletion or inertion. For ingle deletion-correcting code, Levenhtein introduced a redundancy lower bound of log 2 (n) O(1) bit, demontrating that the VT code, which require at mot log 2 (n) redundancy bit, i nearly optimal. The elegance of the VT contruction ha inpired many attempt to extend thi code to correct multiple deletion. Such an approach i found in [7] where the author introduce a number-theoretic contruction that wa later hown in [1] to be capable of correcting two or more deletion. Unfortunately, even for the cae of jut two deletion, the contruction from [7] ha a rate which doe not converge to one. Other contruction for multiple inertion/deletion-correcting code uch a thoe found in [10], [11] rely on (d, k)-contrained code, and conequently, thee code alo have rate le than one. To the bet of the author knowledge, the bet known contruction for two deletion (in term of the minimum redundancy bit) can be found in the recent work by Brakeniek et al. [2]. The author how that it poible to contruct a t deletion-correcting code with c t log 2 n bit of redundancy where c t = O(t 2 log 2 t). The contruction from [2] i for general t and doe not report any improved reult for the cae where t i mall. However, it can be hown that their method reult in a contruction requiring at leat 128 log 2 n bit of redundancy for the cae of t = 2. The bet known lower bound for the redundancy of a double deletion-correcting code i 2 log 2 n O(1) ( [8]) bit; thu, there remain a ignificant gap between the upper and lower bound for t deletion-correcting code even for the cae where t = 2. Thi motivate the effort to earch for more efficient code. We note that uing a counting argument uch a the one found in [9], one can how that there exit a t deletion-correcting code with redundancy at mot 2t log 2 n O(t). However, thee code require the ue of a computer earch to form the codebook along with a lookup table for encoding/decoding. Such code do not cale a n become large and there i no efficient earch mechanim that cale ub-exponentially with n. The contribution of the preent work i a double deletion-correcting code contruction that require 8 log 2 n + O(log 2 log 2 n) bit of redundancy. To the bet of the author knowledge thi repreent the bet contruction for a double deletion-correcting Our Two Deletion Code Code from [2] Redudancy of Code 1200 Bit of Redudancy Code Length Fig. 1. Comparion of new code with code from [2].

2 2 1 Rate of Code Rate Our Two Deletion Code Code from [2] Code Length Fig. 2. Comparion of rate of new code with code from [2]. code in term of the redundancy; it i within a factor of four of the optimal redundancy. In Figure 1 and 2, we compare our contruction and the contruction from [2]. Although both contruction approach rate 1 for long enough code length, our contruction approache rate one much fater than the contruction from [2]. For example, for n = 1024, our contruction ha rate approximately 0.86 wherea the contruction from [2] ha rate at mot The paper i organized a follow. In Section II, we provide the main idea behind our contruction and provide an outline of our approach. Section III introduce our firt contruction. Afterward, an improved contruction i decribed in Section IV. We dicu the iue of run-length-limited contrained code in Section V and conclude with Section VI. II. MAIN IDEAS AND OUTLINE The idea behind our approach i to iolate the deletion of zero and one into eparate equence of information, and to then ue error correction code on the ubtring appearing in the tring. We alo ue a erie of contraint that allow u to detect what type of deletion occurred, and conequently we are able to reduce the number of code in the Hamming metric which are ued a part of the contruction. A a reult, we preent a contruction which achieve the advertied redundancy. Let C(n) F n 2 denote our codebook of length n that i capable of correcting two deletion. Suppoe y F n 2 2 i received where y i the reult of two deletion occurring to ome vector x C(n), denoted y D 2 (x). Then, we have the following 6 cenario: 1) Scenario 1: Two zero were deleted from x. 2) Scenario 2: Two one were deleted from x. 3) Scenario 3: A zero wa deleted from a run of of length 4 and a one wa deleted from a run of length 4 in x. 4) Scenario 4: A ymbol b F 2 wa deleted from a run of length 4 and another ymbol b wa deleted from a run of length l {1, 2, 3}. Furthermore, if l = 1, then b i adjacent to run of length l 1, l 2 where l 1 + l 2 < 4. 5) Scenario 5: A ymbol b F 2 wa deleted from a run of length 4, and a ymbol b i deleted from a run of length 1, where b i adjacent to run of length l1, l 2 where l 1 + l 2 = 4. 6) Scenario 6: Scenario 1)-5) do not occur. The firt 5 of thee cenario are hown in Figure 3. Our approach i to ue a erie of detection code to attempt to delineate between the 6 cenario enumerated above. In addition, and imilar to [2], we make ue of ubtring that are not affected by the deletion. The main difference between our approach here and the one ued by [2] i that we ue a erie of detection code that allow u to place error-correcting code on fewer ubtring occuring in our codeword. We provide an example which illutrate the baic idea behind the approach in [2] and it highlight ome ubequent notation ued throughout the paper. Example 1. Suppoe x = (0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0) i tranmitted. Let L(x, 00) be an integer which denote the maximum number of bit between any two occurrence of the ubtring 00 in x. Notice that L(x, 00) = 4. For horthand, let L(x, 00) = and let f : { } F 1 2 F 2 [2 +1 1] be an injective mapping where denote the null tring. Then, F f (x, 00) = F f4 (x, 00) = (f 4 (0, 1, 1), f 4 ( ), f 4 (1, 0, 1, 1), f 4 ( )). Thu, if there are k occurrence of the ubtring 00, then the equence F f4 (x, 00) ha length k + 1. Suppoe y = (1, 1, 0, 0, 0, 1, 0, 1, 0, 0) i received where y i the reult of two deletion occurring to x. Notice that F f4 (y, 00) = (f 4 (1, 1), f 4 ( ), f 4 (1, 0, 1), f 4 ( )).

3 3 0,1,0,0,0,1,1,1,1,0,1,0,1,0,0 Scenario 1 0,1,0,0,0,1,1,1,1,0,1,0,1,0,0 Scenario 2 0,1,0,0,0,1,1,1,1,0,1,0,1,0,0 Scenario 3 0,1,0,0,0,1,1,1,1,0,1,0,1,0,0 Scenario 4 0,1,0,0,0,1,1,1,1,0,1,0,1,0,0 Scenario 5 Fig. 3. Firt 5 cenario In particular, notice that d H (F f4 (x, 00), F f4 (y, 00)) = 2, where d H denote the Hamming ditance. Thu, if F f4 (x, 00) belong to a double error-correcting code, it i poible to recover F f4 (x, 00) from y and in particular, it i poible to then recover x. Notice in the previou example that all occurrence of the ubtring 00 were preerved (we will more rigorouly define thi notation hortly). It i not too hard to ee that if (in the previou example) a deletion occurred where a ubtring 00 i deleted in x and therefore doe not appear in y, then we can no longer claim d H (F f4 (x, 00), F f4 (y, 00)) = 2. In order to overcome thi iue, the approach taken in [2] wa to require that the equence F f4 (x, w) belong to a double error-correcting code for many different choice of w. In particular, the approach in [2] i to enforce that F f (x, w) hold for every binary tring w of length m where according to Theorem 5 from [2] we need 2 m > 2t (2m 1). Since any two-error correcting code of length n require approximately 2 log 2 n bit of redudancy and t = 2 for our etup, thi implie that uing the contruction from [2], we need m 6 and o the overall contruction require approximately 2 6 (2 log 2 n) = 128 log 2 n bit of redudancy. The approach taken here i to ue a erie of detection code along with different mapping and more carefully chooe which ubtring to place error-correcting code on. Conequently, we how it i poible to contruct a code with fewer redundant bit than the approach outlined in [2] for the cae of two deletion. From the above example if we ue the contraint F f (x, w 1 ), F f (x, w 2 ), F f (x, w 3 ), and F f (x, w 4 ) (for ome appropriately choen ubtring w 1,..., w 4 ), then thi would require the ue of a erie of double error-correcting code defined over an alphabet of ize approximately 2. To reduce the ize of thi alphabet, we make ue of the following lemma. Lemma 1. (c.f., [2]) There i a hah function h : {0, 1} {0, 1} v for v 4 log 2 + O(1), uch that for all x {0, 1}, given any y D 2 (x) and h (x), the tring x can be recovered. Code contructed according to Lemma 1 can be found uing brute force attempt uch a finding an independent et on a graph with 2 n vertice for which no polynomial-time algorithm (with repect to n) exit. Note, however, that if = O(log 2 n), then uing an algorithm for determining a maximal independent et on a graph which i polynomial with repect to the number of vertice in the graph (uch a [6] for intance) reult in a code of length that can be contructed in polynomial time with repect to n. At firt, we will make ue of the equence F h (x, 0000), F h (x, 1111), F h (x, 11011), and F h (x, ) where x C(n). In particular, we will require that each of thee equence belong to a code with minimum Hamming ditance 5 over an alphabet of ize approximately. Auming = O(log 2 n), then thee contraint together require approximately log 2(n) bit of redundancy if we ue the non-binary code from Dumer [5]. Afterward, we alter one of the map ued in conjunction with our Hamming code and how it i poible contruct a code with 8 log 2 n + O(1) bit of redudancy. We now turn to ome additional notation before preenting the contruction. For a vector x F n 2, let D(i 1, i 2, x) F n 2 2 be the reult of deleting the ymbol in x in poition i 1 and i 2 where 1 i 1 < i 2 n. For example if x = (0, 1, 0, 1, 0, 0), then D(2, 4, x) = (0, 0, 0, 0). Uing thi notation, we have D 2 (x) = {y : i 1, i 2, y = D(i 1, i 2, x)}. Let w {0, 1} m. Suppoe y D 2 (x). Then we ay that the ubtring w {0, 1} m i preerved from x to y if, for every occurrence of w, there exit indice i 1 and i 2 uch that D(i 1, i 2, x) = y and the following hold:

4 4 1) w {(x i1, x i1+1,..., x i1+m 1), (x i1 1, x i1,..., x i1+m 2),..., (x i1 m+1, x i1 m+2,..., x i1 )}, 2) w {(x i2, x i2+1,..., x i2+m 1), (x i2 1, x i2,..., x i2+m 2),..., (x i2 m+1, x i2 m+2,..., x i2 )}, 3) w {(y i1 1, y i1,..., y i1+m 2), (y i1 2, y i1 1,..., y i1+m 3),..., (y i1 m+1, y i1 m+2,..., y i1 )}, 4) w {(y i2 2, y i2 1,..., y i2+m 3), (y i2 3, y i2 2,..., y i1+m 4),..., (y i2 m, y i1 m+1,..., y i2 1)}. In word, the firt two tatement above require that any ubtring w i not deleted from x and the lat two tatement require that no new appearance of w are in y that were not alo in x. If w i not preerved from x to y, and the firt two condition above are violated, then we ay that w wa detroyed from x to y. If w i not preerved, and the lat two condition above are violated, then we ay that w wa created from x to y. Notice that in order for w to be preerved from x to y, 1)-4) ha to hold for at leat one pair of i 1, i 2 uch that we can write y = D(i 1, i 2, x) ince the choice of i 1, i 2 may not be unique. The following example how thi. Example 2. Suppoe x = (0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0) and y = (0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0). Then we ay that (1, 0, 1) i preerved from x to y ince there are three occurrence of 101 in x and D(6, 15, x) = D(5, 15, x) = y. Notice that (1, 1, 1) i not preerved from x to y and in particular (1, 1, 1) i detroyed. For a vector x F n 2, let N 0 (x) denote the number of zero in x. Similarly, let N 1 (x) be the number of one that appear in x. Furthermore, let N 0 (x), N 1 (x), N (x) be the number of appearance of the ubtring 0000, 1111, and repectively. We illutrate thee notation in the following example. Example 3. Let x = (0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0). Then, N 0 (x) = 11, N 1 (x) = 10, N 0 = 2, N 1 = 1, and N (x) = 1. Notice that two occurrence of the ubtring 0000 overlap. III. CONSTRUCTION - FIRST ATTEMPT We now turn to decribing out code. Let C T (n, ) denote the following et C T (n, ) = {x F n 2 : L(x, 0000), L(x, 1111), L(x, ), L(x, 11011) }. A we will ee hortly, our main contruction will be a ub-code of C T (n, ). Let c F 5 7. Suppoe q the mallet odd prime greater than the ize of the image of the map h. Suppoe N i the mallet poitive integer uch that q N 1 > n. Let a 0, a 1, a , a F r q where r 2N + N 1 3. Our contruction i the following: C(n,a 0, a 1, a , a 11011, c, ) = { x C T (n, ) : N 0 (x) mod 7 = c 1, N 1 (x) mod 7 = c 2, N 1 (x) mod 7 = c 3, N 0 (x) mod 7 = c 4, N (x) mod 7 = c 5, F h (x, 0000) C 2 (n, q, a 0 ), F h (x, 1111) C 2 (n, q, a 1 ), F h (x, ) C 2 (n, q, a ), } F h (x, 11011) C 2 (n, q, a ), where C 2 (n, q, a) i a code over F q of length n. If any of the equence above that are required to be in code of length n have length M < n, then we imply aume the lat n M component of the equence are equal to zero. Let H be a parity check matrix for a double error-correcting code (minimum Hamming ditance 5) from [5] o that H F r qn 1 q. We define the double error correcting code C 2 (n, q, a) o that C 2 (n, q, a) = {x F qn 1 q : H x = a}. We now how that given any y D 2 (x), it i poible to recover x C(n, a 0, a 1, a , a 11011, c, ). For horthand, we refer to C(n, a 0, a 1, a , a 11011, c, ) a C(n). For the remainder of the ection, we alway aume x i a codeword from C(n) and y D 2 (x). The following claim will be ued throughout the ection. Claim 1. Suppoe a zero i deleted from a run of length one in x and the deletion caue 1) to be created/detroyed, 2) 1111 to be created, then the ubtring i preerved from x to y.

5 5 A an example, the previou claim will be concerned with the following type of deletion: (,,, 1, 0, 1, 1, 0, 1, 1,,,,, ), where indicate a ymbol which i either a zero or a one and 0 repreent a deletion (in thi cae of a ymbol with value 0). Proof: The deletion of a zero from a run of length 1 can detroy a ubtring only if the middle zero i deleted. In thi cae, the ubtring i preerved. The deletion of a zero from a run of length 1 can create a ubtring only if either the firt zero i deleted from the ubtring in x or if the econd zero i deleted from the ubtring in x. In either cae, the ubtring i preerved from x to y. Claim 2. Suppoe a ymbol with value b F 2 i deleted from a run of length 4 and a ymbol with value b i deleted from a run of length 1 where N 1 (x) N 1 (y), N 0 (x) N 0 (y). Under thi etup, if b = 1, the ubtring i preerved. Otherwie, if b = 0 and i not preerved from x to y, then i preerved. For example, the previou claim will be concerned with the following etup. If b = 0, then one intance of the etup from thi claim i and if b = 1, then another example i (,, 0, 0, 0, 0,,,,,, 1, 0, 1,,, ) (,, 1, 1, 1, 1,,,,,, 0, 1, 0,,, ). Proof: Suppoe that a ymbol with value b = 0 i deleted from a run of length 4 and another ymbol with value 0 i deleted from a run of length 1. To begin, notice that the zero which wa deleted from the run of length 4 cannot create/detroy the ubtring 11011, , 1111 from x to y. Then, according to Claim 1, if the deletion of a zero from a run of length 1 a) create/detroy the ubtring and b) create the ubtring 1111 from x to y, then the ubtring i preerved from x to y. Suppoe b = 1. Under thi etup, the ubtring i preerved, and o we can recover x from the contraint F h (x, ) C 2 (n, q, a ). To ee thi, we firt note that an occurrence of the ubtring i detroyed only if a one i deleted from a run of length 2 which i not poible under thi etup. In addition, an occurrence of the ubtring cannot be created by deleting a one from a run of length 1 (ince thi would require that the one i alo adjacent to run of length l 1, l 2 with l 1 + l 2 4 ince a 0000 ubtring i created from x to y) or by deleting a one from a run of length 4. Therefore, i preerved from x to y when b = 1. We begin with the cae where either y i the reult of deleting two zero or two one from x. The firt two lemma handle Scenario 1) and 2) from the previou ection. Lemma 2. Suppoe N 1 (x) N 1 (y) mod 7 = 2. Then, x can be recovered from y. For example, the previou claim will be concerned with the following etup: (,,,, 1,,,,,,,,, 1,,, ). Proof: Since N 1 (x) N 1 (y) mod 7 = 2, two one were deleted from x to obtain y. If N 0 (x) N 0 (y) 0 mod 7, then 0000 i preerved ince the deletion of a 1 can create at mot and o the deletion of two one can create at mot Thu, we conclude that 0 i preerved from x to y. Since two one were deleted, clearly no 0000 ubtring were detroyed. Therefore, 0000 i preerved from x to y, and o we can recover x from y uing the contraint F h (x, 0000) C 2 (n, q, a 0 ). If N 1 (x) N 1 (y) 0 mod 7, then 1111 i preerved, and o we can recover x from y uing the contraint F h (x, 1111) C 2 (n, q, a 1 ). Now we aume that both N 1 (x) N 1 (y) 0 mod 7 and N 0 (x) N 0 (y) 0 mod 7. Note that thi i only poible if a one i deleted from a run of length 4 and a one i deleted from a run of length 1. According to Claim 2, we can determine x from F h (x, ) C 2 (n, q, a ). Next we turn to the cae where two zero have been deleted. Lemma 3. Suppoe N 0 (x) N 0 (y) mod 7 = 2. Then, x can be recovered from y. For example, we will be concerned with the following etup: (,,,, 0,,,,,,,,, 0,,, ).

6 6 Proof: Since N 0 (x) N 0 (y) mod 7 = 2, two zero were deleted from x to obtain y. If N 1 (x) N 1 (y) 0 mod 7, then we can recover x from y uing the contraint F h (x, 1111) C 2 (n, q, a 1 ) uing the ame logic a the previou lemma. In addition, if N 0 (x) N 0 (y) 0 mod 7, then 0000 i preerved, and o we can recover x from y uing the contraint F h (x, 0000) C 2 (n, q, a 0 ). Now we aume that both N 1 (x) N 1 (y) 0 mod 7 and N 0 (x) N 0 (y) 0 mod 7. Similar to the previou lemma, thi i only poible if a zero i deleted from a run of length 4 and a zero i deleted from a run of length 1. We can ue the contraint N (x) mod 7 = c 5 to determine whether the ubtring i preerved from x to y. If i preerved, then we can determine x from F h (x, 11011) C 2 (n, q, a ). If i not preerved then we can determine x from F h (x, ) C 2 (n, q, a ) according to Claim 2. A a reult of the previou two lemma, we aume in the remainder of thi ection that y i the reult of deleting a ymbol with a value 1 and a ymbol with a value 0. The next 3 lemma handle the cae where N 0 (x) N 0 (y) or N 1 (x) N 1 (y). The next lemma cover Scenario 3). Lemma 4. Suppoe N 0 (x) N 0 (y) mod 7 = 1, and N 1 (x) N 1 (y) mod 7 = 1. Then, x can be recovered from y. For example, we will be concerned with the following etup: (,, 0, 0, 0, 0,,,,,,, 1, 1, 1, 1, ). Proof: Since N 0 (x) N 0 (y) mod 7 = 1 and N 1 (x) N 1 (y) mod 7 = 1, y i the reult of deleting a 1 and a 0 from x where both ymbol belong to run of length 4. Since both ymbol were deleted from run of length at leat 4, it follow that no ubtring were created/detroyed and o we can recover x from F h (x, ) C 2 (n, q, a ). The next two lemma handle Scenario 4). Lemma 5. Suppoe N 0 (x) N 0 (y) mod 7 = 1 and N 1 (x) N 1 (y) mod 7 = 0. Then, x can be recovered from y. For example, we will be concerned with the following etup: (,, 0, 0, 0, 0,,,,,,, 0, 1, 1, 0, ). Proof: Since N 0 (x) N 0 (y) mod 7 = 1, clearly a zero wa deleted from a run of zero of length at leat 4 in x. Since N 1 (x) N 1 (y) mod 7 = 0 and there are exactly two deletion (of a zero and a one), no one were deleted from run of one of length 4. Thu, we can recover x from F h (x, 1111) C 2 (n, q, a 1111 ). Lemma 6. Suppoe N 1 (x) N 1 (y) mod 7 = 1 and N 0 (x) N 0 (y) mod 7 = 0. Then, x can be recovered from y. For example, the previou lemma i concerned with the following etup: (,, 1, 1, 1, 1,,,,,,, 1, 0, 0, 1, ). Finally, we turn to the cae where either Scenario 5) or Scenario 6) occur. The next two lemma handle the cae where either N 0 (y) > N 0 (x) or N 1 (y) > N 1 (x). Lemma 7. Suppoe N 0 (y) N 0 (x) mod 7 {1, 2, 3} or N 1 (y) N 1 (x) mod 7 {1, 2, 3}. Then, x can be recovered from y. For example, we will be concerned with the following etup: or (, 0, 0, 0, 1, 0,,,,,,, 1, 0, 0, 1, ), (, 1, 1, 1, 0, 1,,,,,,, 0, 1, 1, 0, ). Proof: Suppoe firt that N 0 (y) N 0 (x) mod 7 {1, 2, 3}. Then, clearly a one from a run of length 1 wa deleted in x reulting in the creation of new 0000 ubtring. If N 1 (y) N 1 (x) mod 7 = 0, then 1111 i preerved from x to y and o we can recover x. Otherwie, if N 0 (y) N 0 (x) mod 7 {1, 2, 3} and N 1 (y) N 1 (x) mod 7 {1, 2, 3}, it follow that a one wa deleted from a run of length 1 and alo a zero wa deleted from a run of length 1, o that we have the following type of etup: (, 0, 0, 0, 1, 0,,,,,, 1, 1, 0, 1, 1, ), The deletion of a one from a run of length 1 cannot or create detroy a ubtring or a ubtring ince the 1 need to be adjacent to two run of length l 1, l 2 where l 1 + l 2 4 ince at leat one 0000 ubtring i created from x to y. Furthermore, the deletion of the one from a run of length 1 clearly cannot create a 1111 ubtring from x to y. Therefore, the deletion of the zero from a run of length 1 create a 1111 ubtring from x to y and from Claim 1, if i not preerved from x to

7 7 y, then i preerved. Thu, we can ue the contraint N (x) mod 7 = c 5, F h (x, ) C 2 (n, q, a ) and F h (x, 11011) C 2 (n, q, a ) to determine x. Notice that it i not poible to have N 0 (y) N 0 (x) mod 7 {1, 2, 3} and N 1 (x) N 1 (y) mod 7 {1, 2, 3}. Thi i becaue in order to have N 0 (y) N 0 (x) mod 7 {1, 2, 3}, a one i deleted from a run of length 1 and thi create a new run of zero of length at leat 4. Then, the deletion of the zero (which alo occur by aumption) can only create 1111 ubtring from x to y and o N 1 (x) N 1 (y) mod 7 {1, 2, 3}. The cae where N 1 (y) N 1 (x) mod 7 {1, 2, 3}, but N 0 (y) N 0 (x) mod 7 {1, 2, 3} can be handled uing the ame logic a before. We have one cae left to conider. Lemma 8. Suppoe N 1 (x) N 1 (y) mod 7 = 0, and N 0 (x) N 0 (y) mod 7 = 0. Then, x can be recovered from y. Proof: Since N 1 (x) N 1 (y) mod 7 = 0 and N 0 (x) N 0 (y) mod 7 = 0, there are two etup to conider: 1) A ymbol b wa deleted from a run of length 4 and another ymbol b from a run of length 1 wa deleted which wa adjacent to a run of length l 1 and another run of length l 2 uch that l 1 + l 2 = 4, o that we have the following type of etup: (, 0, 0, 0, 1, 0,,,,,, 0, 0, 0, 0,, ). 2) The 0000 and 1111 ubtring were preerved from x to y. We firt dicu the decoding procedure and then how that the decoding i correct given that either 1) or 2) hold. Suppoe that N (x) { N (y) mod 7. Then we etimate x to be the equence which agree with at leat two of } the following three contraint: F h (x, 0000) C 2 (n, q, a 0 ), F h (x, 1111) C 2 (n, q, a 1 ), F h (x, 11011) C 2 (n, q, a ). Otherwie, if N (x) N (y) mod 7, we etimate x to be the equence which agree with the contraint F h (x, 0000) C 2 (n, q, a 0 ). Firt uppoe that N (x) N (y) mod 7 and uppoe that 1) hold. If b = 0, then the decoding i correct ince in thi cae x agree with the contraint F h (x, 1111) C 2 (n, q, a 1 ) and F h (x, 11011) C 2 (n, q, a ), ince can be created only if a 0 i deleted from a run of length 1. Notice that if b = 0, then the ubtring cannot be detroyed. Now, uppoe b = 1. In thi cae, if N (x) N (y) mod 7, then the deletion of a zero from { a run of length 1 did not create/detroy the ubtring and o x agree with at leat two of the three contraint from F h (x, 0000) C 2 (n, q, a 0 ), F h (x, 1111) } C 2 (n, q, a 1 ), F h (x, 11011) C 2 (n, q, a ). Thu, the decoding i correct when N (x) N (y) mod 7 and 1) hold. Next conider the { cae where N (x) N (y) mod 7 and uppoe that 2) hold. Then clearly, x agree} with two of the three contraint F h (x, 0000) C 2 (n, q, a 0 ), F h (x, 1111) C 2 (n, q, a 1 ), F h (x, 11011) C 2 (n, q, a ) and o the decoding i correct in thi cae. Suppoe now that N (x) N (y) mod 7 and that 1) hold. Notice that under thi etup, b 0, ince the deletion of a 0 from a run of length at leat 4 and the deletion of a 1 from a run of length 1 cannot create/detroy any ubtring from x to y. If b = 1 and N (x) N (y) mod 7, then a zero wa deleted from a run of length one and a one wa deleted from a run of length 4 o that the ubtring 0000 wa preerved and o the decoding i correct in thi cae. Finally, we conider the cae where N (x) N (y) mod7 and 2) hold. If 2) hold and the 0000 ubtring i preerved then clearly x agree with the contraint F h (x, 0000) C 2 (n, q, a 0 ), and o the decoding procedure i correct in thi cae a well. A a conequence of Lemma 2-8, we have the following theorem. Theorem 9. The code C(n, a 0, a 1, a , a 11011, c, ) can correct two deletion. In the next ection, we make ome modification to the code dicued in thi ection and afterward we dicu the redudancy of the reulting code. IV. AN IMPROVED CONSTRUCTION In thi ection, we modify the contruction in the previou ection to obtain a code with redudancy 8 log 2 n + O(log 2 log 2 n). Our contruction ue the ame ubtring to partition our codeword a in the previou ection, but we make ue of a different hah function in place of h from Lemma 1, denoted h (R). Conequently we how that we can replace the contraint F h (x, 11011) C 2 (n, q, a ) with the contraint that F (R) h (x, 11011) belong to a code with Hamming ditance 3 (rather than Hamming ditance 5). Our analyi and the ubequent proof will mirror the previou ection in light of thee modification. Thi ection i organized a follow. We firt decribe our code contruction in detail and then how it ha the advertied redudancy. Afterward, we prove the code can correct two deletion.

8 8 Let C T 2 (n, ) denote the following et where for a binary vector v, τ(v) i the length of the longet run of of zeroe or one in v, C T 2 (n, ) = {x F n 2 : L(x, 0000), L(x, 1111), L(x, ), L(x, 11011), τ(x) }. In the following, for a vector v F n 2, let r 1 (x) denote the run-length repreentation of the run of one in x. For example, if v = (1, 1, 0, 1, 0, 1, 1, 1), then r 1 (v) = (2, 1, 3). Furthermore, let r 2 be the run-length repreentation of one in x with length at leat 2. For example, r 2 (v) = (2, 3). Let Q be the mallet prime greater than. We now turn to decribing the map h (R) : F 2 F 2 Q. Let H R1 F 2 Q be the parity check matrix for a code C L with Hamming ditance at leat 3 over F Q. For a vector v {0, 1} we define h (R) a the vector which reult by conidering the run-length repreentation (a a vector) of the run of one in v of length at leat 2 and multiplying H R1 by thi vector. We provide an example of thi map next. Example 4. Suppoe v = (0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0). Then, the vector repreenting the run of one in v i r 1 (v) = (2, 1, 2). Notice that r 1 (v) ha an alphabet ize which i equal to the length of the longet run in v. Then, h (R) (v) = H R1 r 2 (v) = H R1 (2, 2). Let c F 6 7. Suppoe q 1 the mallet odd prime greater than the ize of the image of the map h, and let q 2 be the mallet prime greater than the ize of the image of the map h (R). A before, let N 1 be the mallet poitive integer uch that q N1 1 1 > n, and uppoe N 2 i the mallet poitive integer uch that q N2 2 1 > n. Let a 0, a 1, a F r1 q 1 and let a F r2 q 2 where r 1 2N 1 + N1 1 3 and r N 2. In the following, let b Z +1. Our contruction i the following: C (2) (n,a 0, a 1, a , a 11011, c, b, ) = N 0 (x) mod 7 = c 1, N 1 (x) mod 7 = c 2, N 0 (x) mod 7 = c 3, N 1 (x) mod 7 = c 4, { x C T 2 (n, ) : N (x) mod 7 = c 5, N (x) mod 7 = c 6 F h (x, 0000) C 2 (n, q 1, a 0 ), F h (x, 1111) C 2 (n, q 1, a 1 ), F h (x, ) C 2 (n, q 1, a ), F (R) h (x, 11011) C 1 (n, q 2, a ), } r 1 (x) i = b mod + 1, where C 1 (n, q 2, a ) i either a primitive BCH code with root {1, α} (α F q N 2 2 i odd i an element of order q N2 2 1) or a coet of uch a code. If any of the equence above that are required to be in code of length n have length M < n, then we imply aume the lat n M component of the equence are equal to zero. Since the parameter c, a 0, a 1, a , a 11011, and b can be choen arbitrarily, it follow uing an averaging argument that there exit a choice of c, a 0, a 1, a , a 11011, and b that give C (2) (n,a 0, a 1, a , a 11011, c, b, ) C T 2 (n, ) 7 6 q 3r1 1 q r2 2 ( + 1). Auming that the image of the map h ha cardinality 2 4 log 2 () and = 128 log 2 (n) then we can approximate q 1 = (128 log 2 n) 4. In addition, if q N1 1 log 1 = n + 1, then N 1 = 2 (n+1) 4 log 2 (128 log 2 (n)) + 1. Then, r N 1, and o log 2 q r1 1 = 7 3 log 2(n + 1) + O(log 2 log 2 n). Auming Q = + 1 = 128 log 2 (n) + 1, then we approximtae q 2 = 256 log 2 (n) + 2. In addition if q N2 2 = n + 2, then N 2 = log 2 (n+2) log 2 (256 log 2 (n)+2). Since r N 2, we have log q r2 2 log 2(n + 2) + O(log 2 log 2 (n)).

9 9 Thu, log 2 C (2) (n,a 0, a 1, a , a 11011, c, b, ) log 2 C T 2 (n, ) 8 log n O(log 2 log 2 n). In the next ection, we how that when 128 log 2 (n), log 2 C T 2 (n, ) n O(1) and o there exit a code which meet our lower bound. We now prove that the code C (2) (n, a 0, a 1, a , a 11011, c, b, ) can correct two deletion by conidering the ame cenario a in the previou ection. With a light abue of notation, in thi ection C(n) will denote C (2) (n, a 0, a 1, a , a 11011, c, b, ) and NOT C(n, a 0, a 1, a , a 11011, c, ) from the previou ection. Analagou to the previou ection, we begin with the following claim and throughout we aume x C(n). Claim 3. Suppoe a zero i deleted from a run of length one in x and the deletion caue 1) to be created/detroyed, 2) 1111 to be created, then the ubtring i preerved from x to y, and d H (F h (R) ) (x, 11011), F (R) h (y, 11011) = 1. Proof: The deletion of a zero from a run of length 1 can detroy a ubtring only if one of the middle zero i deleted. In thi cae, the ubtring 1111 cannot be created from x to y. The only other cae to conider i when a ubtring i created and alo the ubtring 1111 i created. Notice that thi i only poible when the firt zero i deleted from the ubtring or if the lat zero i deleted from the ubtring Notice that under either etup, the ubtring ) i preerved. In addition r 2 ( ) = (3, 2) and r 2 ( ) = (4, 2) o that d H (F (R) h (notice r 2 ( ) = (2, 3) and r 2 ( ) = (2, 4) ). (x, 11011), F h (R) (y, 11011) Claim 4. Suppoe that y i the reult of deleting a zero in x from a run of length 1 where the zero i adjacent to run of length 1 and length l where l 3. Then, given r 2 (x), i odd r 1(x) i mod + 1, and y, it i poible to determine r 1 (x). Proof: Since y i the reult of deleting a zero from a run of length 1 where the zero i adjacent to run of length 1 and l 3, it follow that r 2 (y) can be obtained by ubtituting a ymbol in r 2 (y) which ha value l with another ymbol which ha value l + 1. Since l + 1 3, it follow that d H (r 2 (y), r 2 (x)) = 1 and o we can determine the location of the ymbol in r 2 (y) that wa altered a a reult of the deletion to y. To obtain r 1 (x) from r 1 (y), r 2 (x), r 2 2(y), we replace the ymbol, ay a, that ha value l + 1 in r 1 (y) which correpond to the ame ymbol in r 2 (x) (that wa affected by the deletion of the zero in x) and we replace the ymbol a in r 1 (y) with 2 adjacent ymbol 1 and l. Given i odd r 1(x) i mod + 1, we can determine whether the ymbol 1 hould be inerted before the ymbol l or whether l come before the ymbol 1. Thu, we can recover r 1 (x) a tated in the claim. We have the following lemma which mirror the logic from the previou ection. The firt lemma follow immediately uing the ame logic a in the proof of Lemma 2. Lemma 10. Suppoe N 1 (x) N 1 (y) mod 7 = 2. Then, x can be recovered from y. The next lemma require a little more work. Lemma 11. Suppoe N 0 (x) N 0 (y) mod 7 = 2. Then, x can be recovered from y. Proof: Similar to the proof of Lemma 3, we focu on the cae where both N 1 (x) N 1 (y) 0 mod 7 and N 0 (x) N 0 (y) 0 mod 7, ince if at mot one of thee two condition hold then we can determine x from y given F h (x, 0000) C 2 (n, q, a 0 ), F h (x, 1111) C 2 (n, q, a 1 ). Since N 1 (x) N 1 (y) 0 mod 7 and N 0 (x) N 0 (y) 0 mod 7 hold, it follow that y i the reult of deleting a zero from a run of length 1 and another zero from a run of length at leat 4. Notice that the deletion of the zero from a run of length 4 cannot create/detroy the ubtring , Thu, if the ubtring i not preerved from x to y, it i a reult) of the deletion of the zero from a run of length 1. According to Claim 3, under thi etup, d H (F h (R) Thu, we can determine r 2 (x) from F h (R) (x, 11011), F (R) h (y, 11011) = 1. (x, 11011) C 1 (n, q 2, a ). According to Claim 4, we can then determine r 1 (x) o that we can correct the deletion of the zero from a run of length 1. The remaining deletion (of a zero from a run of length 4) can be corrected uing the contraint F h (x, 1111) C 2 (n, q 1, a 2 ). Thu, we can determine x from y a follow. Suppoe N 1 (x) N 1 (y) 0 mod 7, N 0 (x) N 0 (y) 0 mod 7, and N (x) N (y) mod 7. Then, x can be recovered a decribed in the previou paragraph uing the contraint F h (R) C 1 (n, q 2, a ), F h (x, 1111) C 2 (n, q 1, a 1 ). Otherwie, if N 1 (x) N 1 (y) 0 mod 7, N 0 (x) N 0 (y) 0 mod 7, and N (x) N (y) mod 7, x can be recovered from y uing F h (x, ) C 2 (n, q 1, a ). = 1 (x, 11011)

10 10 The next lemma can be proven in the ame manner a Lemma 4. Lemma 12. Suppoe N 0 (x) N 0 (y) mod 7 = 1, and N 1 (x) N 1 (y) mod 7 = 1. Then, x can be recovered from y. The proof of the next two lemma are the ame a Lemma 5 and Lemma 6. Lemma 13. Suppoe N 0 (x) N 0 (y) mod 7 = 1 and N 1 (x) N 1 (y) mod 7 = 0. Then, x can be recovered from y. Lemma 14. Suppoe N 1 (x) N 1 (y) mod 7 = 1 and N 0 (x) N 0 (y) mod 7 = 0. Then, x can be recovered from y. Next, we conider the cae where either N 0 (y) > N 0 (x) or N 1 (y) > N 1 (x). The reult can be proven uing idea imilar to Lemma 7 and Lemma 11. The proof can be found in Appendix A. Lemma 15. Suppoe N 0 (y) N 0 (x) mod 7 {1, 2, 3} or N 1 (y) N 1 (x) mod 7 {1, 2, 3}. Then, x can be recovered from y. The next lemma follow uing imilar idea a from Lemma 8. The proof can be found in Appendix B. Lemma 16. Suppoe N 1 (x) N 1 (y) mod 7 = 0, and N 0 (x) N 0 (y) mod 7 = 0. Then, x can be recovered from y. V. CONSTRAINT REDUNDANCY The purpoe of thi ection i to how that there i no aymptotic rate lo incurred by tarting with our contrained equence pace where there are no more than ymbol between conecutive appearance of v 1 = 0000, v 2 = 1111, v 3 = 11011, v 4 = , v 5 = 1 and v 6 = 0. Our goal will be to how that the probability a equence of length n atifie thee contraint converge to a contant for n. Let E i be the event that contraint i i met by ome equence a elected uniformly at random from {0, 1} n. We how that for all ufficiently large n, 6 i=1 P r(e i) > 5. Then, we have by union bound P r(e 1 E 6 ) = 1 P r(ē1 Ē6) 6 1 P r(ēi) > 0. (1) Thi implie that the redundancy incurred i indeed a contant number of bit. To compute E i, we ue the following argument. Set X i to be a random variable counting the number of appearance of the contraint equence v i in a. Let Y i be the number of ymbol between two conecutive appearance of the contraint. Then, conditioning on X i, i=1 P r(e i ) = x P r(y i ) x P r(x i = x) = E[P r(y i ) Xi ]. The next tep i to ue the fact that c x i convex and apply Jenen inequality: E[P r(y i ) Xi ] P r(y i ) E[Xi]. Firt, we can eaily compute the E[X i ] term. Let u et M = v i to be the contraint equence length. In our cae, M {1, 4, 5, 6}. Let a j i = (a i, a i+1,..., a j ) be a ubequence of a. Then, uing linearity of expectation, we have that E[X i ] = n M+1 i=1 P r(a j i = v i) = (n M + 1)/2 M. Next, we conider the P r(y i ) probabilitie; thee probabilitie meaure waiting time. A ueful obervation i that contraint equence uch a have lower waiting time compared to the equal-length pattern Thi can be eaily formalized through a Markov chain analyi: we model the appearance of a contraint equence through a erie of tate forming a Markov chain. In the cae of , the appearance of any 0 prior to the final tate take u back to the initial tate. However, thi i not the cae for (or any other mixed-run equence of the ame length), ince, among other example, the appearance of a 1 in tate 11 take u back to the 11 tate rather than the initial tate. Therefore, ince we are lower bounding our probability, we may directly work with the contraint equence 1111, , and The 0000 contraint i identical to Next, oberve that the waiting time atify the tochatic recurion Y M+1 = Y M +1+BỸM+1, where B i a Bernoulli random variable with parameter 1/2, all variable are independent, and ỸM+1 i ditributed like Y M+1. It can be hown that the olution to thi recurion are exponential random variable; more pecifically, 2 M Y M converge (in ditribution) to an exponential r.v. with parameter 1/2. Then, we may write, for ufficiently large n, P r(y i ) = P r(2 M Y i 2 M ) 1 exp( 2 (M+1) ).

11 11 Thu, for contraint i with length M, the probability P r(e i ) i lower bounded a ( ( P r(e i ) 1 exp )) n M+1 2 M 2 M+1. We can afely ignore the term in the exponent that are not n/2 M, ince the reulting expreion goe to 1. Next, we ubtitute the value of. Since = 2 M+1 log 2 (n), we have that ( ( 1 exp )) n 2 M 2 M+1 = (1 exp ( log 2 (n))) n 2 M ( = 1 1 ) n 2 M n ) (1 n 2 M 2 M = = e 2 M. For our contraint, we have M = (4, 4, 5, 6, 1, 1), yielding lower bound on 6 i=1 P r(e i) of o that the condition in (1) i met. VI. CONCLUSION In thi work, we provided a contruction for a code capable of correcting two deletion improved upon exiting art in term of the number of redundant bit. Open problem include extending thee technique to code capable of correcting multiple deletion a well a developing imple encoding technique. REFERENCES [1] K.A.S. Abdel-Ghaffar, F. Paluncic, H.C. Ferreira, and W.A. Clarke, On Helberg generalization of the Levenhtein code for multiple deletion/inertion error correction, IEEE Tranaction on Information Theory, vol. 58, no. 3, pp , [2] J. Brakeniek, V. Guruwami, and S. Zbarky, Efficient low-redundancy code for correcting multiple deletion, in Proceeding of the Twenty-Seventh Annual ACM-SIAM Sympoium on Dicrete Algorithm, pp , SIAM, [3] S. Datta and S. W. McLaughlin, An enumerative method for runlength-limited code: permutation code, IEEE Tranaction on Information Theory, vol. 45, no. 6, pp , [4] S. Datta and S. W. McLaughlin, Optimal block code for M-ary runlength-contrained channel, IEEE Tranaction on Information Theory, vol. 47, no. 5, pp , [5] I. Dumer, Nonbinary double-error-correcting code deigned by mean of algebraic varietie, IEEE Tran. Inf. Theory, vol. 41, no. 6, pp , Nov [6] F. V. Fomin, F. Grandoni, and D. Kratch, A meaure & conquer approach for the analyi of exact algorithm, Journal of the ACM (JACM), vol. 56, no. 5, pp. 25, 2009 [7] A.S.J. Helberg and H.C. Ferreira, On multiple inertion/deletion correcting code, IEEE Tranaction on Information Theory, vol. 48, no. 1, pp , [8] A.A. Kulkarni, and N. Kiyavah, Nonaymptotic upper bound for deletion correcting code, IEEE Tranaction on Information Theory, vol. 59, no. 8, pp , [9] V.I. Levenhtein, Binary code capable of correcting deletion, inertion, and reveral Soviet Phyic-Doklady, vol. 10, no. 8, pp , [10] F. Paluncic, Khaled A.S. Abdel-Ghaffar, H.C. Ferreira, and W.A. Clarke, A multiple inertion/deletion correcting code for run-length limited equence, IEEE Tranaction on Information Theory, vol. 58 no. 3, pp , [11] R. M. Roth and P.H. Siegel, Lee-metric BCH code and their application to contrained and partial-repone channel, vol. 40, no. 4, pp , [12] F. F. Seller, Jr., Bit lo and gain correction code, IRE Tranaction on Information Theory, vol. 8, no. 1, pp , [13] R. R. Varhamov and G. M. Tenengolt, A code for correcting a ingle aymmetric error, Avtomatika i Telemekhanika, vol. 26, no. 2, pp , n 2 M APPENDIX A PROOF OF LEMMA 15 Lemma 15. Suppoe N 0 (y) N 0 (x) mod 7 {1, 2, 3} or N 1 (y) N 1 (x) mod 7 {1, 2, 3}. Then, x can be recovered from y. Proof: We conider the cae where N 0 (y) N 0 (x) mod 7 {1, 2, 3} and N 1 (y) N 1 (x) mod 7 {1, 2, 3} ince otherwie x can be recovered from y uing the ame idea a in the proof of Lemma 7. The deletion of a one from a run of length 1 cannot create/detroy a ubtring or a ubtring ince the 1 need to be adjacent to two run of length l 1, l 2 where l 1 + l 2 4 ince at leat one 0000 ubtring i created from x to y. Furthermore, the deletion of the one from a run of length 1 clearly cannot create a 1111 ubtring from x to y. Therefore, the deletion of the zero from a run of length 1 create a 1111 ubtring from x to y. If, in addition, the deletion of the zero doe alo doe not preerve the ubtring from ) x (y, 11011) to y, then according to Claim 3, the ubtring i preerved from x to y and d H (F (R) h (x, 11011), F (R) h Uing the ame logic a in the proof of Lemma 11, according to Claim 4 we can correct the deletion of a zero from a run of length 1 uing the contraint F (R) h (x, 11011) C 1 (n, q 2, a ) and i odd r 1(x) i = b mod + 1 and the deletion of the one from a run of length 4 can be corrected with the contraint F h (x, 0000) C 2 (n, q 1, a 0 ). Thu, the decoding in thi cae i the ame a decribed in the lat paragraph of Lemma 11. = 1.

12 12 APPENDIX B PROOF OF LEMMA 16 Lemma 16. Suppoe N 1 (x) N 1 (y) mod 7 = 0, and N 0 (x) N 0 (y) mod 7 = 0. Then, x can be recovered from y. Proof: Uing the ame logic a before, N 1 (x) N 1 (y) mod 7 = 0 and N 0 (x) N 0 (y) mod 7 = 0, there are two etup to conider: 1) A ymbol b wa deleted from a run of length 4 and another ymbol b from a run of length 1 wa deleted which wa adjacent to a run of length l 1 and another run of length l 2 uch that l 1 + l 2 = 4. 2) The 0000 and 1111 ubtring were preerved from x to y. The decoding procedure i the following. Suppoe N {(x) N (y) mod 7. Then we etimate x to be the equence which agree with at leat two of the following three contraint: F h (x, 0000) C 2 (n, q 1, a 0 ), F h (x, 1111) C 2 (n, q 1, a 1 ), F h (x, ) } C 2 (n, q 1, a ). Otherwie, if N (x) N (y) mod 7, we etimate x to be the equence which agree with the contraint F h (x, 0000) C 2 (n, q 1, a 0 ). Firt uppoe that N (x) N (y) mod 7 and uppoe that 1) hold. If b = 0, then the decoding i correct ince in thi cae x agree with the contraint F h (x, 1111) C 2 (n, q 1, a 2 ) and F h (x, ) C 2 (n, q 1, a ), ince it i not poible to create a 0000 ubtring and alo to create/detroy Now, uppoe b = 1. In thi cae, if N (x) N (y) mod 7, then the deletion of a zero from { a run of length 1 did not create/detroy the ubtring and o x agree with at leat two of the three contraint from F h (x, 0000) C 2 (n, q 1, a 0 ), F h (x, 1111) C 2 (n, q 1, a 1 ), F h (x, ) } C 2 (n, q 1, a ). Thu, the decoding i correct when N (x) N (y) mod 7 and 1) hold. Next conider the cae { where N (x) N (y) mod 7 and uppoe that 2) hold. Then clearly, x agree with } two of the three contraint F h (x, 0000) C 2 (n, q 1, a 0 ), F h (x, 1111) C 2 (n, q 1, a 1 ), F h (x, ) C 2 (n, q 1, a ) and o the decoding i correct in thi cae. Suppoe now that N (x) N (y) mod 7 and that 1) hold. Notice that under thi etup, b 0, ince the deletion of a 0 from a run of length at leat 4 and the deletion of a 1 from a run of length 1 that create a 0000 ubtring cannot create/detroy any ubtring from x to y. If b = 1 and N (x) N (y) mod 7, then a zero wa deleted from a run of length one and a one wa deleted from a run of length 4 o that the ubtring 0000 wa preerved and o the decoding i correct in thi cae. Finally, we conider the cae where N (x) N (y) mod7 and 2) hold. If 2) hold and the 0000 ubtring i preerved then clearly x agree with the contraint F h (x, 0000) C 2 (n, q 1, a 0 ), and o the decoding procedure i correct in thi cae a well.

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